A solution is prepared by adding 1.60 g of solid NaCl to 50.0 mL of 0.100 mol L^-1 CaCl₂. What is the molarity of chloride ion in the final solution? Assume that the volume of the final solution is 50.0 mL.

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Quizplus · Accepted Answer

The moles of Cl⁻ from CaCl₂ are 0.100 mol/L * 0.050 L * 2 = 0.010 mol. The moles of Cl⁻ from NaCl are 1.60 g / 58.44 g/mol = 0.0274 mol. Total moles of Cl⁻ = 0.010 + 0.0274 = 0.0374 mol. Molarity = 0.0374 mol / 0.050 L = 0.748 mol/L, which rounds to 0.747 mol/L.

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